Skip to main content

Buffer Maker (Henderson–Hasselbalch)

Design and adjust buffer recipes from stock solutions or dry reagents. Compute volumes to reach target pH, plot buffer/titration curves, and estimate buffer capacity.

Loading calculator...

Last Updated: November 18, 2025. This content is regularly reviewed to ensure accuracy and alignment with current buffer chemistry principles.

Understanding Buffer Solutions and the Henderson–Hasselbalch Equation

Buffer solutions are the unsung heroes of chemistry and biology—they resist dramatic pH changes when small amounts of acid or base are added, maintaining stable pH ranges essential for countless chemical reactions, biological processes, and analytical procedures. A buffer consists of a conjugate acid-base pair: a weak acid (HA) and its conjugate base (A⁻), or a weak base (B) and its conjugate acid (BH⁺), existing together in equilibrium. When acid (H⁺) is added, the conjugate base neutralizes it; when base (OH⁻) is added, the weak acid neutralizes it—the pH shifts minimally because the buffer system "absorbs" these perturbations. This pH-stabilizing ability makes buffers indispensable in biological systems (blood pH 7.35-7.45 is buffered by carbonate, phosphate, and protein systems), enzyme assays (where pH affects activity), cell culture (maintaining optimal growth conditions), and analytical chemistry (controlling reaction conditions).

The Henderson–Hasselbalch equation—pH = pKa + log₁₀([A⁻]/[HA])—is the fundamental tool for understanding and designing buffer systems. Derived from the acid dissociation equilibrium (Ka = [H⁺][A⁻] / [HA]) by taking negative logarithms and rearranging, this elegant formula connects three key quantities: pH (the acidity you want), pKa (the characteristic acid strength of your buffer system, which you choose), and [A⁻]/[HA] (the ratio of conjugate base to acid, which you control by mixing appropriate amounts). The beauty of Henderson–Hasselbalch is its simplicity: it tells you exactly what ratio is needed to achieve a target pH, or conversely, what pH results from a given ratio. For students, mastering this equation unlocks buffer problems on exams, homework, and lab reports.

Conjugate acid-base pairs are central to buffer chemistry. If you start with a weak acid like acetic acid (CH₃COOH, pKa = 4.76), its conjugate base is acetate (CH₃COO⁻). Mix acetic acid with sodium acetate (which dissociates completely to provide acetate ions), and you have an acetate buffer. The weak acid can donate protons to neutralize added base: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The conjugate base can accept protons to neutralize added acid: CH₃COO⁻ + H⁺ → CH₃COOH. Because both forms are present in significant quantities (neither fully dissociated nor fully protonated), the system can counteract pH changes in both directions. This equilibrium-based resistance is what makes buffers work.

This calculator is a comprehensive tool for student-level buffer planning and conceptual understanding. It supports multiple modes: (1) Calculate pH from pKa and ratio—given a buffer system (pKa) and the ratio [A⁻]/[HA], instantly find the resulting pH using Henderson–Hasselbalch. (2) Find required ratio for target pH—if you need a buffer at pH 5.0 and you're using a pKa 4.76 system (acetic acid), the calculator tells you the ratio needed (e.g., [A⁻]/[HA] = 1.74). (3) Design buffer recipes from stock solutions—input stock concentrations of acid and base, desired final volume, and target pH, and the calculator computes conceptual volumes of each stock to mix (for understanding, not actual lab preparation). (4) Estimate buffer capacity—qualitatively understand how well your buffer resists pH change based on total concentration and pH relative to pKa.

Real-world buffer applications are vast. In biochemistry, enzyme activity is pH-sensitive—trypsin works at pH 8, pepsin at pH 2—so buffers maintain optimal conditions throughout experiments. In molecular biology, Tris-HCl buffers (pKa ≈ 8.1) are standard for DNA/RNA work; phosphate-buffered saline (PBS, pH 7.4) mimics physiological conditions. In analytical chemistry, buffers control pH in titrations, chromatography mobile phases, and colorimetric assays. In cell culture, HEPES and bicarbonate buffers maintain pH 7.2-7.4 despite CO₂ fluctuations. In medicine, intravenous fluids are buffered to match blood pH. For students, understanding buffers is essential for general chemistry, analytical chemistry, biochemistry, and lab courses—buffer calculations appear on virtually every chemistry exam and are fundamental to lab technique.

Critical safety and scope disclaimer: This calculator is designed for educational, homework, and conceptual learning purposes only. It helps you understand the Henderson–Hasselbalch equation, practice buffer pH calculations, compare buffer systems, and build chemical intuition. It does NOT provide instructions for actual laboratory buffer preparation, which requires proper training, calibrated equipment (analytical balances, pH meters, volumetric glassware), chemical safety protocols, and adherence to standard operating procedures. Never use this tool to prepare solutions for biological experiments, medical applications, food/beverage production, or any context where pH accuracy is critical for safety or function. Real-world buffer preparation involves considerations beyond this calculator's scope: purity of reagents, ionic strength, temperature effects, sterility (for biological use), and verification with calibrated pH meters. Use this calculator to learn the theory—consult trained professionals, validated protocols, and proper equipment for practical buffer work.

Understanding the Basics of Buffer Systems

What is a Buffer Solution?

A buffer solution is an aqueous mixture that resists changes in pH when small amounts of acid (H⁺) or base (OH⁻) are added. Buffers contain significant concentrations of both a weak acid (HA) and its conjugate base (A⁻), or a weak base (B) and its conjugate acid (BH⁺). The weak acid can neutralize added base: HA + OH⁻ → A⁻ + H₂O. The conjugate base can neutralize added acid: A⁻ + H⁺ → HA. Because both forms are present in substantial amounts (the weak acid is only partially dissociated), the buffer can counteract pH changes in either direction, maintaining relatively stable pH. Without a buffer, adding even small amounts of strong acid or base would cause large pH swings—buffers "smooth out" these perturbations through their equilibrium-based resistance.

What are Conjugate Acid-Base Pairs?

A conjugate acid-base pair consists of two species that differ by a single proton (H⁺). When a weak acid HA loses a proton, it becomes its conjugate base A⁻. Conversely, when A⁻ gains a proton, it becomes HA. Example: acetic acid (CH₃COOH) and acetate ion (CH₃COO⁻) are conjugates. Ammonia (NH₃, a weak base) and ammonium ion (NH₄⁺, its conjugate acid) are another pair. In a buffer, you deliberately create a mixture containing both members of the conjugate pair—this is what enables buffering. If you only had the weak acid with no conjugate base, adding base would consume all the acid, drastically changing pH. If you only had the base with no acid, adding acid would consume all the base, also changing pH drastically. Having both present means the system can absorb additions from either direction.

What is pKa and Why Does It Matter for Buffers?

pKa is the negative base-10 logarithm of the acid dissociation constant (pKa = -log₁₀Ka). It measures the strength of a weak acid: lower pKa = stronger acid (more prone to donating protons). For buffer design, pKa is crucial because buffers work best when pH is close to pKa. Specifically, effective buffering occurs within pH = pKa ± 1 (called the buffer range). At pH = pKa, the ratio [A⁻]/[HA] = 1 (equal amounts of acid and base), which provides maximum buffer capacity. Outside pKa ± 1, either the acid or base form dominates, leaving little of the other to neutralize additions—buffering becomes weak. Therefore, to design a buffer for pH 5, choose an acid with pKa ≈ 4-6. Common buffer systems: acetate (pKa 4.76) for pH 3.8-5.8, phosphate (pKa₂ 7.2) for pH 6.2-8.2, Tris (pKa 8.1) for pH 7-9, ammonia (pKa 9.25) for pH 8.25-10.25.

The Henderson–Hasselbalch Equation Explained

The Henderson–Hasselbalch (HH) equation is:

pH = pKa + log₁₀([A⁻] / [HA])

Where [A⁻] is the concentration of conjugate base (M) and [HA] is the concentration of weak acid (M). This equation is derived from the acid dissociation equilibrium: Ka = [H⁺][A⁻] / [HA]. Taking -log₁₀ of both sides and rearranging gives the HH form. The equation tells us: (1) If [A⁻] = [HA] (ratio = 1), then log₁₀(1) = 0, so pH = pKa (the midpoint of the buffer range). (2) If [A⁻] > [HA] (more base than acid), log is positive, so pH > pKa (buffer is on the basic side). (3) If [A⁻] < [HA] (more acid than base), log is negative, so pH < pKa (buffer is on the acidic side). The beauty of HH is that pH depends on the ratio, not absolute concentrations—doubling both [A⁻] and [HA] doesn't change pH (though it increases buffer capacity).

Buffer Range: Why pH Should Be Near pKa

The effective buffer range is pH = pKa ± 1. Within this range, the ratio [A⁻]/[HA] is between 1:10 and 10:1—both forms are present in significant amounts, enabling effective neutralization of added acid or base. At pH = pKa - 1, ratio = 10⁻¹ = 0.1 (10× more acid than base, but base is still 10% of total). At pH = pKa + 1, ratio = 10¹ = 10 (10× more base than acid, but acid is still 9% of total). Outside this range, one form becomes negligible: at pH = pKa + 2, ratio = 100:1 (only 1% acid)—adding acid overwhelms the small amount of weak acid present, and pH drops sharply. The ± 1 rule ensures both forms remain in at least 10% abundance, providing robust buffering. This is why you can't use an acetate buffer (pKa 4.76) to buffer pH 8—you'd be 3 pH units away, ratio would be 10³ = 1000, meaning 99.9% base and 0.1% acid—no acid left to neutralize added base.

What is Buffer Capacity?

Buffer capacity (β) quantifies how much acid or base a buffer can neutralize before pH changes significantly. It's maximized when: (1) pH = pKa (equal amounts of acid and base), and (2) total concentration is high (more moles of buffer to absorb additions). Buffer capacity is roughly β ≈ 2.303 × [HA] × [A⁻] / ([HA] + [A⁻]). For equal concentrations (pH = pKa), this simplifies to β ≈ 0.576 × C_total. Practically: a 0.1 M acetate buffer at pH 4.76 has higher capacity than a 0.01 M buffer at the same pH—the former can absorb 10× more acid/base before pH shifts 1 unit. Similarly, a buffer at pH = pKa has higher capacity than one at pH = pKa + 1 (where one form dominates). This explains why biochemists use concentrated buffers (e.g., 50-100 mM) for experiments—higher capacity ensures pH stability despite metabolic acids/bases produced during reactions.

Why Ratio, Not Absolute Amounts, Determines pH

A profound insight from Henderson–Hasselbalch: pH depends on the ratio [A⁻]/[HA], not the individual concentrations. A buffer with 0.1 M acetate and 0.1 M acetic acid (ratio = 1) has the same pH as one with 0.01 M acetate and 0.01 M acetic acid (ratio = 1)—both give pH = pKa = 4.76. This is because pH is determined by the relative proportions of acid and base forms, which control the equilibrium position of HA ⇌ H⁺ + A⁻. However, buffer capacity differs: the 0.1 M buffer (10× more concentrated) can neutralize 10× more added acid/base before the ratio shifts significantly. Practical implication: diluting a buffer with pure water doesn't change its pH (ratio stays constant) but reduces its capacity to resist pH changes. Conversely, adding equal amounts of acid and base to a buffer increases capacity without changing pH.

Common Buffer Systems and Their Applications

Acetate buffers (CH₃COOH/CH₃COO⁻, pKa 4.76): pH 3.8-5.8, used in enzyme assays (e.g., lysozyme), protein electrophoresis, and acidic extraction procedures. Phosphate buffers (H₂PO₄⁻/HPO₄²⁻, pKa₂ 7.2): pH 6.2-8.2, versatile for near-neutral pH work, protein purification, and HPLC mobile phases. Tris buffers (Tris-HCl, pKa 8.1): pH 7-9, standard in molecular biology for DNA/RNA work, Western blots, and PCR. HEPES (pKa 7.5): pH 6.8-8.2, "Good's buffer" for cell culture (minimal toxicity, doesn't chelate metals). Carbonate buffers (H₂CO₃/HCO₃⁻, pKa₁ 6.4): physiological buffering in blood. Citrate buffers (pKa 3.1, 4.8, 6.4): multiprotic, used in food science and clinical chemistry. Ammonia buffers (NH₃/NH₄⁺, pKa 9.25): pH 8.25-10.25, for basic pH work. Choosing the right buffer means matching pKa to your target pH range.

How to Use the Buffer Maker (Henderson–Hasselbalch) Calculator

This calculator supports multiple modes for buffer calculations. Each mode addresses a different aspect of buffer design and analysis. Here's a comprehensive guide:

Mode 1: Calculate pH from pKa and Ratio (Basic Henderson–Hasselbalch)

Use this when you know your buffer system (pKa) and the ratio of conjugate base to acid, and you want to find the resulting pH.

Step 1: Enter the pKa Value

Input the pKa of your weak acid buffer system. For acetic acid, pKa = 4.76. For phosphate, use pKa₂ = 7.20. For Tris, pKa ≈ 8.06.

Step 2: Enter Concentrations or Ratio

Input either: (a) [A⁻] and [HA] concentrations (e.g., 0.15 M acetate and 0.10 M acetic acid), or (b) the ratio [A⁻]/[HA] directly (e.g., 1.5). The calculator will compute the ratio if you give concentrations.

Step 3: Calculate pH

Click Calculate. The tool applies pH = pKa + log₁₀(ratio) and returns the buffer pH. It may also show whether pH is within the effective buffer range (pKa ± 1).

Step 4: Interpret Results

Check if pH is near pKa (good buffering) or far from pKa (weak buffering). Verify that pH makes sense: if [A⁻] > [HA], pH should be > pKa. If [A⁻] < [HA], pH should be < pKa.

Example: Acetate buffer with pKa = 4.76, [A⁻] = 0.2 M, [HA] = 0.1 M.

Input: pKa = 4.76, [acetate] = 0.2, [acetic acid] = 0.1

Output: ratio = 2, pH = 4.76 + log₁₀(2) = 4.76 + 0.30 = 5.06

Interpretation: pH > pKa because more base than acid.

Mode 2: Find Required Ratio for Target pH

Use this when you have a target pH in mind and want to know what ratio [A⁻]/[HA] is needed.

Step 1: Enter pKa and Target pH

Input the pKa of your buffer system and the pH you want to achieve. For example, pKa = 4.76 (acetate) and target pH = 5.00.

Step 2: Calculate Required Ratio

The calculator rearranges Henderson–Hasselbalch: [A⁻]/[HA] = 10^(pH - pKa). For pH = 5.00 and pKa = 4.76, ratio = 10^(0.24) ≈ 1.74. This means you need 1.74 times as much conjugate base as acid.

Step 3: Check Feasibility

If target pH is outside pKa ± 1, the calculator may warn that buffering will be weak. Ratios > 10 or < 0.1 indicate you're at the edge of the effective range.

Example: Need pH 7.4 using phosphate buffer (pKa 7.2).

Input: pKa = 7.2, target pH = 7.4

Output: ratio = 10^(7.4 - 7.2) = 10^0.2 ≈ 1.58

Interpretation: Need 1.58× more HPO₄²⁻ than H₂PO₄⁻.

Mode 3: Design Buffer from Stock Solutions (Conceptual Recipe)

Use this to understand how to mix stock solutions of acid and base to achieve a target pH and volume. For educational understanding only—not actual lab instructions.

Step 1: Enter Stock Concentrations

Input the concentration of your acid stock (e.g., 1 M acetic acid) and base stock (e.g., 1 M sodium acetate).

Step 2: Enter Target pH and Final Volume

Specify desired pH (e.g., 5.0) and total buffer volume (e.g., 100 mL). Also input desired final concentration of buffer components.

Step 3: Calculate Conceptual Volumes

The calculator determines volumes of acid stock, base stock, and water to mix. It uses: (a) Henderson–Hasselbalch to find ratio, (b) dilution equation (C₁V₁ = C₂V₂) to find individual volumes, (c) mass/volume balance to ensure total = final volume.

Step 4: Understand Limitations

These are conceptual numbers for learning. Real buffer preparation requires pH meter verification, consideration of ionic strength, temperature effects, and proper lab technique. Never use calculator outputs directly in lab without validation.

Mode 4: Adjust Existing Buffer pH (Conceptual)

Use this to understand how adding strong acid or base affects an existing buffer's pH.

Step 1: Enter Current Buffer Composition

Input current pH, volume, pKa, and concentrations of acid and base forms.

Step 2: Enter Target pH

Specify the new pH you want to reach (within pKa ± 1 ideally).

Step 3: Calculate Required Addition

The calculator estimates how much strong acid (e.g., HCl) or strong base (e.g., NaOH) to add. This is conceptual—real pH adjustment requires incremental additions with pH meter monitoring.

Universal Tips for All Modes

  • Always verify that target pH is within pKa ± 1 for effective buffering.
  • Remember that pH depends on ratio, but buffer capacity depends on total concentration.
  • Use log₁₀ (common logarithm), not natural log (ln), in all Henderson–Hasselbalch calculations.
  • Check that results make sense: if you increase [A⁻] relative to [HA], pH should increase.
  • Read calculator warnings—they indicate when you're outside effective buffer range or when pipetting volumes are impractical.
  • All outputs are for conceptual understanding and homework practice, not actual laboratory procedures.

Formulas and Mathematical Logic Behind Buffer Calculations

Understanding the mathematics empowers you to solve buffer problems on exams, verify calculator results, and build intuition about buffer behavior.

1. The Henderson–Hasselbalch Equation (Core Formula)

pH = pKa + log₁₀([A⁻] / [HA])

[A⁻]: Molar concentration of conjugate base (M)
[HA]: Molar concentration of weak acid (M)
pKa: Negative logarithm of acid dissociation constant (dimensionless)
log₁₀: Common (base-10) logarithm, NOT natural log

Key insights: When [A⁻] = [HA], ratio = 1, log₁₀(1) = 0, so pH = pKa. When [A⁻] > [HA], ratio > 1, log is positive, so pH > pKa. When [A⁻] < [HA], ratio < 1, log is negative, so pH < pKa.

2. Rearranged Forms of Henderson–Hasselbalch

Solve for ratio from pH and pKa:

[A⁻] / [HA] = 10^(pH - pKa)

Solve for pKa from pH and ratio:

pKa = pH - log₁₀([A⁻] / [HA])

These rearrangements are useful when solving for different unknowns. For example, if you know pH and pKa, use the first form to find the required ratio. If you know pH and ratio, use the second to back-calculate pKa.

3. Buffer Capacity Formula (Simplified)

β ≈ 2.303 × [HA] × [A⁻] / ([HA] + [A⁻])

β: Buffer capacity (mol/(L·pH unit))
Maximum when [HA] = [A⁻] (pH = pKa): β_max ≈ 0.576 × C_total

Buffer capacity quantifies how many moles of strong acid or base can be added per liter before pH changes by 1 unit. It's maximized at pH = pKa and increases with total buffer concentration.

4. Calculating Individual Concentrations from Ratio and Total Concentration

If you know the ratio R = [A⁻]/[HA] and total buffer concentration C_total = [HA] + [A⁻], you can find individual concentrations:

[A⁻] = C_total × R / (1 + R)

[HA] = C_total × 1 / (1 + R)

Example: C_total = 0.1 M, R = 2
[A⁻] = 0.1 × 2 / 3 = 0.0667 M
[HA] = 0.1 × 1 / 3 = 0.0333 M

5. Dilution and Buffer pH

Key principle: Diluting a buffer with pure water does NOT change its pH (ideally), because the ratio [A⁻]/[HA] remains constant—both concentrations decrease proportionally. However, buffer capacity decreases proportionally to dilution.

Example: 100 mL of buffer with [A⁻] = 0.2 M, [HA] = 0.1 M.

Initial ratio = 0.2 / 0.1 = 2, pH = pKa + 0.30

Dilute to 200 mL (2× dilution):

[A⁻] = 0.1 M, [HA] = 0.05 M

New ratio = 0.1 / 0.05 = 2 (unchanged)

pH still = pKa + 0.30 (same!)

But capacity is now half (concentrations halved).

6. Worked Example: Find pH from pKa and Ratio

Problem: Acetate buffer with pKa = 4.76. [Acetate] = 0.15 M, [Acetic acid] = 0.10 M. Find pH.

Step 1: Calculate ratio

Ratio = [A⁻] / [HA] = 0.15 / 0.10 = 1.5

Step 2: Apply Henderson–Hasselbalch

pH = pKa + log₁₀(ratio)

pH = 4.76 + log₁₀(1.5)

pH = 4.76 + 0.176 = 4.94

Answer:

pH = 4.94 (slightly above pKa because more base than acid)

7. Worked Example: Find Required Ratio for Target pH

Problem: Need phosphate buffer at pH 7.4. Phosphate pKa₂ = 7.20. What ratio [HPO₄²⁻]/[H₂PO₄⁻] is needed?

Step 1: Rearrange Henderson–Hasselbalch

[A⁻] / [HA] = 10^(pH - pKa)

Step 2: Plug in values

Ratio = 10^(7.4 - 7.2) = 10^0.2

Ratio = 1.585

Answer:

Need 1.585× as much HPO₄²⁻ as H₂PO₄⁻. For example, 0.158 M HPO₄²⁻ and 0.100 M H₂PO₄⁻.

8. Worked Example: Effect of Adding Strong Base to Buffer

Problem (Conceptual): 1 L of acetate buffer: [Acetate] = 0.1 M, [Acetic acid] = 0.1 M, pKa = 4.76. Add 0.01 mol NaOH. What's new pH?

Step 1: Initial pH

Ratio = 0.1 / 0.1 = 1, pH = 4.76 + log₁₀(1) = 4.76

Step 2: Reaction with buffer

OH⁻ reacts with HA: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

0.01 mol HA consumed, 0.01 mol A⁻ produced

New [HA] = 0.1 - 0.01 = 0.09 M

New [A⁻] = 0.1 + 0.01 = 0.11 M

Step 3: New pH

New ratio = 0.11 / 0.09 = 1.222

pH = 4.76 + log₁₀(1.222) = 4.76 + 0.087 = 4.85

Answer:

pH changes from 4.76 to 4.85 (only 0.09 pH units!). Without buffer, 0.01 M NaOH in 1 L would give pH 12. This demonstrates buffer resistance.

Practical Applications of Buffer Calculations

Buffer calculations are essential for student understanding across chemistry coursework. Here are detailed student-focused scenarios (all conceptual, not actual lab procedures):

1. Homework Problem: Finding Buffer pH from Composition

Scenario: Your general chemistry homework asks: "Calculate the pH of a buffer containing 0.25 M NH₃ and 0.40 M NH₄Cl. pKa of NH₄⁺ = 9.25." You recognize this as a Henderson–Hasselbalch problem. NH₃ is the base (conjugate base of NH₄⁺), NH₄⁺ is the acid. Ratio = [NH₃]/[NH₄⁺] = 0.25/0.40 = 0.625. pH = 9.25 + log₁₀(0.625) = 9.25 + (-0.204) = 9.05. The calculator confirms your answer instantly, letting you check your work and build confidence before moving to the next problem. You learn: buffer pH is below pKa when there's more acid than base.

2. Exam Question: Choosing the Right Buffer for a Target pH

Scenario: An exam asks: "You need a buffer at pH 6.5. Which system is best: (A) acetic acid/acetate (pKa 4.76), (B) phosphate H₂PO₄⁻/HPO₄²⁻ (pKa 7.20), or (C) ammonia/ammonium (pKa 9.25)?" You know buffers work within pKa ± 1. Option A: pKa 4.76, range 3.76-5.76 (too acidic). Option C: pKa 9.25, range 8.25-10.25 (too basic). Option B: pKa 7.20, range 6.20-8.20 (includes 6.5!). Answer: B. The calculator lets you verify by computing pH 6.5 for each system—option B requires a reasonable ratio (0.63), while A and C require extreme ratios (> 50 or < 0.01), confirming weak buffering.

3. Lab Report: Understanding Buffer Capacity Conceptually

Scenario: Your analytical chemistry lab report asks: "Why does a 0.1 M acetate buffer resist pH change better than a 0.01 M acetate buffer at the same pH?" You understand from buffer capacity formula (β ∝ C_total) that higher concentration = higher capacity. The 0.1 M buffer has 10× more moles of buffer per liter, so it can absorb 10× more added acid/base before pH shifts 1 unit. Use the calculator to explore: compute how much strong acid you'd need to drop pH by 0.5 units for each buffer—you'd need ~10× more for the concentrated buffer, demonstrating capacity. This builds intuition about concentration vs. pH.

4. Problem Set: Effect of Adding Strong Acid to Buffer

Scenario: Problem: "A buffer contains 0.2 M acetate and 0.2 M acetic acid in 500 mL. Add 0.005 mol HCl. Find new pH (pKa 4.76)." Initial: ratio = 1, pH = 4.76. HCl (strong acid) dissociates completely, H⁺ reacts with acetate: CH₃COO⁻ + H⁺ → CH₃COOH. Moles of A⁻ decrease by 0.005, moles of HA increase by 0.005. New [A⁻] = (0.2 × 0.5 - 0.005) / 0.5 = 0.19 M. New [HA] = (0.2 × 0.5 + 0.005) / 0.5 = 0.21 M. New ratio = 0.19/0.21 = 0.905. pH = 4.76 + log(0.905) = 4.72. pH drops only 0.04 units—buffer resists! The calculator walks you through these steps, reinforcing the reaction stoichiometry.

5. Biochemistry Context: Understanding Physiological Buffers

Scenario: Your biochemistry homework asks: "Blood is buffered at pH 7.4 by H₂CO₃/HCO₃⁻ (pKa 6.4). What is the ratio [HCO₃⁻]/[H₂CO₃] in blood?" Use rearranged HH: ratio = 10^(7.4 - 6.4) = 10^1 = 10. So blood has 10× more bicarbonate than carbonic acid. This explains why blood can neutralize dietary acids (HCO₃⁻ accepts H⁺) but is also why hyperventilation (blowing off CO₂, which forms H₂CO₃) increases pH (respiratory alkalosis). The calculator makes this abstract concept concrete—you see the ratio that maintains life.

6. Molecular Biology Application: Tris Buffer pH at Different Temperatures

Scenario (Conceptual): Problem: "Tris buffer (pKa 8.06 at 25°C) has strong temperature dependence: pKa decreases by ~0.03 per °C. If you prepare Tris at pH 8.0 at 25°C, what's the pH at 37°C (body temperature)?" At 37°C, pKa ≈ 8.06 - (0.03 × 12) = 7.70. The ratio [Tris]/[TrisH⁺] doesn't change with temperature (same moles), but pKa does. pH = pKa + log(ratio). If initial pH = pKa = 8.06 at 25°C (ratio = 1), then at 37°C, pH ≈ 7.70 (ratio still 1). The buffer's pH drops 0.36 units! This teaches why you must prepare and use buffers at the same temperature, especially for Tris.

7. Analytical Chemistry: Buffer Selection for HPLC Mobile Phase

Scenario (Conceptual): You're studying HPLC in analytical chemistry. Problem: "Select a buffer for pH 3.0 mobile phase. Options: citrate (pKa₁ 3.1), acetate (pKa 4.76), phosphate (pKa₂ 7.2)." For pH 3.0, you need pKa ≈ 2-4. Citrate pKa₁ = 3.1 is within range (3.1 ± 1 = 2.1-4.1). Acetate's range is 3.76-5.76 (pH 3.0 is below, ratio would be 0.17, weak buffering). Phosphate is far off. Answer: citrate. The calculator confirms by showing citrate at pH 3.0 requires ratio [citrate²⁻]/[H₂-citrate⁻] = 10^(3.0 - 3.1) = 0.79 (reasonable), while acetate requires 0.17 (edge of range).

8. Advanced Problem: Buffer Dilution and Capacity Trade-offs

Scenario: Problem: "You have 100 mL of 0.5 M phosphate buffer at pH 7.4. You dilute to 1 L. Does pH change? Does capacity change?" From theory: dilution doesn't change pH (ratio [HPO₄²⁻]/[H₂PO₄⁻] stays constant—both concentrations decrease 10×). But capacity decreases 10× because β ∝ C_total. Use calculator to verify: initial C_total = 0.5 M (high capacity), final C_total = 0.05 M (1/10 capacity). pH remains 7.4 in both cases. This demonstrates a key principle: you can't increase capacity by dilution, only by adding more buffer salts. Practical lesson: for experiments needing strong buffering, use concentrated buffers, not diluted ones.

Common Mistakes in Buffer Calculations

Buffer problems are error-prone due to logarithms, ratios, and equilibrium concepts. Here are the most frequent mistakes and how to avoid them:

1. Using Concentrations Instead of Ratio in Henderson–Hasselbalch

Mistake: Calculating pH = pKa + log₁₀([A⁻]) without dividing by [HA].

Why it's wrong: Henderson–Hasselbalch requires the ratio [A⁻]/[HA], not just [A⁻]. If [A⁻] = 0.2 M and [HA] = 0.1 M, you must compute 0.2/0.1 = 2, then pH = pKa + log₁₀(2) = pKa + 0.30. Just using 0.2 gives pH = pKa + log₁₀(0.2) = pKa - 0.70 (completely wrong).

Solution: Always divide [conjugate base] by [acid] before taking the log. The ratio is what matters for pH, not individual concentrations.

2. Confusing pKa with Ka in Formulas

Mistake: Plugging pKa (e.g., 4.76) directly into Ka expressions, or vice versa.

Why it's wrong: pKa = -log₁₀(Ka), so Ka = 10⁻ᵖᴷᵃ. If pKa = 4.76, then Ka = 10⁻⁴·⁷⁶ ≈ 1.74 × 10⁻⁵, not 4.76. Using pKa value as if it were Ka leads to wildly incorrect results. Henderson–Hasselbalch uses pKa, while equilibrium expressions (Ka = [H⁺][A⁻]/[HA]) use Ka.

Solution: Check formula requirements: Henderson–Hasselbalch needs pKa (logarithmic form). Equilibrium constant expressions need Ka (exponential form). Convert if necessary.

3. Using Natural Log (ln) Instead of Common Log (log₁₀)

Mistake: Calculating pH = pKa + ln([A⁻]/[HA]) instead of using log₁₀.

Why it's wrong: Henderson–Hasselbalch is defined with base-10 logarithm (log₁₀), not natural logarithm (ln). If ratio = 2, log₁₀(2) = 0.301, but ln(2) = 0.693. Using ln inflates pH change by factor of 2.303, giving completely wrong pH.

Solution: Always use log₁₀ (or "log" button on calculator, which defaults to base 10). Never use ln for pH calculations.

4. Applying Henderson–Hasselbalch to Strong Acids/Bases

Mistake: Trying to use HH equation for HCl/NaCl or NaOH/NaCl mixtures.

Why it's wrong: Henderson–Hasselbalch applies to weak acid/conjugate base systems in equilibrium. Strong acids (HCl, HNO₃) dissociate completely—no equilibrium exists, so HH doesn't apply. Mixing HCl with NaCl doesn't create a buffer; it's just a strong acid solution with spectator ions.

Solution: Only use HH for weak acid/conjugate base (or weak base/conjugate acid) pairs. For strong acids/bases, calculate pH directly from [H⁺] or [OH⁻].

5. Forgetting to Account for Stoichiometry When Adding Acid/Base

Mistake: When strong acid/base is added to a buffer, not updating [HA] and [A⁻] before recalculating pH.

Why it's wrong: Adding H⁺ reacts with A⁻ to form HA (A⁻ decreases, HA increases by the same amount). Adding OH⁻ reacts with HA to form A⁻ (HA decreases, A⁻ increases). If you use original concentrations in HH, you ignore this reaction, giving wrong pH.

Solution: First, determine how much HA and A⁻ change due to the reaction (stoichiometry: 1:1 for strong acid/base with buffer). Then use new concentrations in HH to find new pH.

6. Using pH Outside the Buffer Range (pKa ± 1)

Mistake: Designing a buffer at pH 8 using acetic acid (pKa 4.76), which is 3.24 pH units away.

Why it's wrong: At pH 8, the ratio [acetate]/[acetic acid] = 10^(8 - 4.76) = 10³·²⁴ ≈ 1738. This means 99.94% acetate, 0.06% acetic acid—essentially no acid left to neutralize added base. Buffering is negligible.

Solution: Always choose a buffer system with pKa within ± 1 of target pH. If pKa is far from pH, select a different buffer system. The calculator will warn you if ratio is extreme.

7. Mixing Units Inconsistently (Molarity, mM, moles)

Mistake: Using [A⁻] = 100 mM and [HA] = 0.05 M in ratio without converting.

Why it's wrong: Ratio = 100 mM / 0.05 M = 100 / 50 = 2 (if correctly converted: 100 mM = 0.1 M, so 0.1/0.05 = 2). But if you forget to convert, you might compute 100/0.05 = 2000 (wrong by 1000×).

Solution: Convert all concentrations to the same unit (typically Molarity, M) before computing ratio. Alternatively, use the raw amounts (moles) since ratio of moles = ratio of concentrations (if in same volume).

8. Assuming Dilution Changes Buffer pH

Mistake: Thinking that diluting a buffer 10× will change its pH.

Why it's wrong: Dilution with pure water decreases both [A⁻] and [HA] by the same factor, leaving ratio [A⁻]/[HA] unchanged. Since pH depends on ratio, pH stays constant (ideally). However, buffer capacity decreases proportionally.

Solution: Remember: pH from ratio (doesn't change with dilution), capacity from concentration (does change). Dilution reduces capacity but not pH (unless dilution is extreme and assumptions break down).

9. Not Checking if Resulting Ratio is Physically Reasonable

Mistake: Getting ratio = 0.001 or ratio = 1000 and not questioning it.

Why it's wrong: Extreme ratios indicate you're far outside the buffer range or made a calculation error. Ratio < 0.1 or > 10 means one form dominates (< 10% of the other), so buffering is weak.

Solution: Sanity check: for effective buffering, ratio should be between 0.1 and 10 (pH within pKa ± 1). If you get extreme ratio, verify: (a) pKa is appropriate for target pH, (b) you didn't mix up pKa and Ka, (c) you used log₁₀ not ln.

10. Forgetting That Henderson–Hasselbalch is an Approximation

Mistake: Expecting HH to give exact pH in all conditions.

Why it's wrong: HH assumes: (a) activities ≈ concentrations (valid in dilute solutions), (b) no side reactions, (c) complete dissociation of salts. At high ionic strength, activities deviate. For very dilute buffers (< 0.001 M), water auto-ionization matters. At extreme pH, assumptions break.

Solution: HH is excellent for homework/exam problems under typical conditions (0.01-1 M, pH 3-11). For research-level accuracy, use activity coefficients or rigorous equilibrium calculations. For student work, HH is appropriate.

Advanced Tips for Mastering Buffer Calculations

Once you've mastered basics, these advanced strategies deepen understanding and prepare you for complex buffer chemistry:

1. Always Match pKa to Target pH for Optimal Buffering

Strategy: When designing a buffer, choose a system whose pKa is as close as possible to your target pH (ideally within 0.5 pH units). Maximum buffer capacity occurs at pH = pKa, and capacity drops off rapidly as you move away. For pH 7.4 (physiological), phosphate (pKa 7.2) or HEPES (pKa 7.5) are excellent; acetate (pKa 4.76) would be terrible. Build a mental pKa library: acetate ~4.8, phosphate ~7.2, Tris ~8.1, ammonia ~9.3, carbonate ~10.3. This lets you quickly select appropriate systems on exams or in lab planning.

2. Understand Buffer Capacity Quantitatively, Not Just Qualitatively

Insight: Buffer capacity β = 2.303 × [HA] × [A⁻] / ([HA] + [A⁻]) tells you exactly how many moles of strong acid/base can be added per liter to shift pH by 1 unit. For equal concentrations ([HA] = [A⁻] = C/2), β ≈ 0.576 × C. So a 0.1 M buffer at pH = pKa has β ≈ 0.058 mol/(L·pH). Adding 0.058 mol/L of acid/base shifts pH by 1 unit. This quantitative understanding lets you design buffers with specific capacity requirements, not just hoping "high concentration = good."

3. Exploit the Log Relationship: Small Ratio Changes = Small pH Changes

Mathematical insight: Because pH = pKa + log₁₀(ratio), the logarithm "dampens" ratio changes. Doubling the ratio (ratio 1 → 2) only changes pH by 0.30 units. Halving (2 → 1) also changes pH by 0.30. This is why buffers work—large absolute concentration changes translate to small pH changes through the logarithm. Conversely, unbuffered solutions experience huge pH shifts with small additions because there's no buffering equilibrium. Understanding this log relationship builds intuition about buffer resistance.

4. Recognize When to Use Multiprotic Buffer Systems

Advanced application: Multiprotic acids (H₃PO₄, citric acid, carbonic acid) have multiple pKa values, each corresponding to a buffering region. Phosphate has pKa₁ = 2.15, pKa₂ = 7.20, pKa₃ = 12.35—it can buffer in three ranges: pH 1-3 (H₃PO₄/H₂PO₄⁻), pH 6-8 (H₂PO₄⁻/HPO₄²⁻), pH 11-13 (HPO₄²⁻/PO₄³⁻). The most useful is pKa₂ for near-neutral pH. Citrate (pKa₁ 3.1, pKa₂ 4.8, pKa₃ 6.4) provides broad buffering across acidic pH. Understanding which pKa dominates in your pH range is crucial for multiprotic buffers.

5. Master Mental Approximations for Quick pH Estimates

Exam technique: Memorize log₁₀ values: log(1) = 0, log(2) ≈ 0.3, log(5) ≈ 0.7, log(10) = 1. If ratio = 2, pH = pKa + 0.3 (easy mental math). If ratio = 0.5 = 1/2, log(0.5) = -log(2) ≈ -0.3, so pH = pKa - 0.3. For ratio = 3, log(3) ≈ 0.48 ≈ 0.5. These approximations let you estimate buffer pH in seconds on multiple-choice exams, eliminating wrong answers without calculator. Practice with powers of 2, 5, and 10.

6. Understand Temperature Effects on Buffer pKa and pH

Thermodynamic connection: pKa changes with temperature due to ΔH° of dissociation. Tris buffer has strong temperature dependence (dpKa/dT ≈ -0.03 per °C), while phosphate is more stable (dpKa/dT ≈ -0.003). If pKa changes but ratio [A⁻]/[HA] stays constant (same moles), pH will change. Practical: prepare buffers at the temperature you'll use them. For Tris at 4°C (cold room), pKa ≈ 8.8; at 37°C (incubator), pKa ≈ 7.7—a 1.1 pH unit shift! Phosphate is preferred for temperature-variable work.

7. Connect Buffer Behavior to Le Chatelier's Principle

Conceptual framework: Buffers work via Le Chatelier. Equilibrium: HA ⇌ H⁺ + A⁻. Add H⁺ → equilibrium shifts left (H⁺ + A⁻ → HA), consuming added H⁺, minimizing pH drop. Add OH⁻ → OH⁻ reacts with H⁺ (H⁺ + OH⁻ → H₂O), removing H⁺ → equilibrium shifts right (HA → H⁺ + A⁻) to replenish H⁺, minimizing pH rise. Buffer capacity is the extent of this equilibrium shift, determined by amounts of HA and A⁻ present. Viewing buffers through Le Chatelier provides deep qualitative understanding alongside the quantitative HH equation.

8. Appreciate the Role of Ionic Strength in Real Buffers

Advanced consideration: In concentrated solutions, ionic strength affects activity coefficients (γ), so pH = -log₁₀(γ_H × [H⁺]) deviates from simple pH = -log₁₀([H⁺]). High ionic strength (from buffer salts, NaCl, etc.) decreases activity coefficients, slightly shifting pH from HH predictions. For homework, ignore this (use concentrations). For research, use activity corrections or empirical pH measurement. Understanding this shows the limits of HH and why pH meters are essential for real buffer work.

9. Explore Buffer Mixtures and Overlapping Ranges

Sophisticated application: Some protocols use mixtures of buffers with different pKa values to achieve broad-range pH stability or specific capacity profiles. Example: "universal buffers" combine citrate (pKa 3-6), phosphate (pKa 7), and borate (pKa 9) to cover pH 2-12 with single stock. While not needed for homework, understanding multi-buffer systems prepares you for complex analytical/biological applications. Each component buffers in its range; total capacity is the sum.

10. Build Your Personal Buffer pKa Reference Table

Study strategy: Memorize pKa values and applications of 8-10 common buffers: acetate (4.76, enzyme assays), citrate (3.1/4.8/6.4, acidic work), phosphate (7.2, versatile), Tris (8.1, molecular biology), HEPES (7.5, cell culture), MES (6.1, slightly acidic), MOPS (7.2, near-neutral), ammonia (9.25, basic). Knowing these by heart lets you select appropriate buffers instantly on exams, design buffers mentally in lab planning, and understand biochemical protocols when they specify "50 mM Tris pH 8" (you know: Tris pKa 8.1, so pH 8 is near maximum capacity). Flashcards work well.

Limitations & Assumptions

• Henderson-Hasselbalch Approximations: The Henderson-Hasselbalch equation (pH = pKa + log[A⁻]/[HA]) assumes activity coefficients equal 1 and that equilibrium concentrations equal initial concentrations. These assumptions fail for dilute buffers, extreme pH values, or high ionic strength solutions.

• Effective pH Range Limited: Buffers only work effectively within approximately ±1 pH unit of the pKa. Outside this range, buffer capacity drops dramatically. Using a buffer far from its pKa provides minimal pH stabilization and poor resistance to acid/base addition.

• Temperature Dependence: pKa values change with temperature. Biological buffers like Tris have significant temperature coefficients (ΔpKa/°C ≈ -0.03). A buffer prepared at room temperature may have significantly different pH at 4°C or 37°C. Always specify and control temperature for critical applications.

• Ionic Strength Effects: High ionic strength (from buffer salts, NaCl, etc.) affects activity coefficients and shifts apparent pKa values. For precise biochemical work, use activity corrections or empirically determine pH under actual experimental conditions.

Important Note: This calculator is strictly for educational and informational purposes only. It demonstrates buffer chemistry principles for learning and homework verification. For biochemistry research, cell culture, or analytical applications, verify buffer pH with a calibrated pH meter under actual experimental conditions.

Sources & References

The buffer chemistry principles and Henderson-Hasselbalch calculations referenced in this content are based on authoritative chemistry sources:

pKa values are temperature-dependent. Values listed are typically for 25°C (298 K). Adjust buffer pH for temperature variations in critical applications.

Frequently Asked Questions

What is a buffer solution in simple terms?
A buffer solution is a mixture that resists changes in pH when small amounts of acid or base are added. It contains a weak acid (like acetic acid) and its conjugate base (like acetate ion), or a weak base and its conjugate acid. When acid (H⁺) is added, the conjugate base neutralizes it. When base (OH⁻) is added, the weak acid neutralizes it. This dual-action resistance maintains relatively stable pH, which is crucial for enzyme activity, cell function, and chemical reactions that are pH-sensitive. For example, blood is buffered at pH 7.4 by carbonate, phosphate, and protein systems—without buffering, eating acidic foods would drastically shift blood pH.
How does the Henderson–Hasselbalch equation work?
The Henderson–Hasselbalch equation, pH = pKa + log₁₀([A⁻]/[HA]), connects three key buffer properties: pH (the acidity you want), pKa (the characteristic strength of your weak acid, which you choose), and [A⁻]/[HA] (the ratio of conjugate base to acid, which you control). Derived from the acid dissociation equilibrium by taking logarithms and rearranging, it tells you: (1) if base and acid are equal ([A⁻] = [HA]), then pH = pKa, (2) if more base than acid, pH &gt; pKa, (3) if more acid than base, pH &lt; pKa. The beauty is that pH depends on the ratio, not absolute amounts—doubling both concentrations doesn't change pH but increases buffer capacity. This equation is your main tool for buffer design and analysis.
Why must I choose a buffer with pKa close to my target pH?
Buffers work best when pH is within pKa ± 1 (the effective buffer range). This is because the ratio [A⁻]/[HA] needs to be between 0.1 and 10 (1:10 to 10:1) for both forms to exist in significant amounts—if ratio &lt; 0.1 or &gt; 10, one form dominates and the other is nearly absent, leaving little to neutralize added acid or base. For example, if you need pH 5, choose a buffer with pKa 4-6. Acetate (pKa 4.76) is perfect—at pH 5, ratio = 10^(5-4.76) = 1.74 (reasonable). But Tris (pKa 8.1) is terrible—at pH 5, ratio = 10^(5-8.1) = 0.00079 (99.9% acid, 0.1% base, no buffering). Matching pKa to pH ensures robust buffering.
What is the difference between pH and pKa in buffer calculations?
pH is the actual acidity of your solution (pH = -log₁₀[H⁺]), which changes based on what's in the solution. pKa is the characteristic acid strength of a specific weak acid (pKa = -log₁₀Ka), an intrinsic property that doesn't change (except with temperature). In buffers, pH is what you want to achieve, and pKa is what you're given by your choice of buffer system. The Henderson–Hasselbalch equation connects them: pH = pKa + log₁₀(ratio). If pH = pKa, the ratio is 1 (equal amounts). If pH &gt; pKa, you need more base than acid. If pH &lt; pKa, you need more acid than base. Think of pKa as the 'anchor point' around which your buffer works—pH can vary within pKa ± 1.
Can this calculator help me prepare real lab buffers?
This calculator is for educational, homework, and conceptual understanding only—not actual laboratory buffer preparation. It helps you learn Henderson–Hasselbalch, practice buffer pH calculations, compare buffer systems, and build chemical intuition. Real buffer preparation requires: (1) calibrated pH meters for verification, (2) analytical balances for precise masses, (3) volumetric glassware for accurate dilutions, (4) high-purity reagents, (5) consideration of ionic strength, temperature effects, and sterility (for biological use), (6) adherence to validated protocols and safety procedures. Never use calculator outputs directly in lab, clinical, food, or safety-critical applications. Use this tool to understand buffer theory—consult trained professionals and proper equipment for practical buffer work.
What does the ratio [A⁻]/[HA] actually mean in Henderson–Hasselbalch?
The ratio [A⁻]/[HA] is the relative proportions of conjugate base to weak acid in your buffer, expressed as concentrations (M). It determines pH through Henderson–Hasselbalch. Ratio = 1 means equal amounts (pH = pKa). Ratio = 2 means twice as much base as acid (pH = pKa + 0.30). Ratio = 0.5 means half as much base as acid (pH = pKa - 0.30). Ratio = 10 means 10× more base (pH = pKa + 1, edge of buffer range). The ratio is what changes when you add strong acid/base: adding H⁺ converts A⁻ to HA (ratio decreases, pH drops); adding OH⁻ converts HA to A⁻ (ratio increases, pH rises). Because pH depends on ratio, not absolute concentrations, diluting a buffer doesn't change pH (ratio stays constant) but reduces capacity.
Does dilution change buffer pH or capacity?
Dilution with pure water does NOT change buffer pH (ideally) because the ratio [A⁻]/[HA] remains constant—both concentrations decrease by the same factor, leaving their ratio unchanged. Since pH = pKa + log₁₀(ratio), and ratio is constant, pH stays the same. However, buffer capacity DOES decrease proportionally with dilution because capacity depends on total concentration (β ≈ 0.576 × C_total at pH = pKa). If you dilute a buffer 10×, pH remains the same but capacity drops 10×—it can absorb only 1/10 as much acid/base before pH shifts significantly. Practical lesson: to increase capacity, add more buffer salts, don't just dilute less. For experiments needing strong buffering, use concentrated buffers.
How do I calculate buffer pH when I know the pKa and ratio?
Use the Henderson–Hasselbalch equation directly: pH = pKa + log₁₀([A⁻]/[HA]). Step 1: Find or calculate the ratio. If given concentrations, divide [conjugate base] / [weak acid]. If given moles, divide moles_base / moles_acid (in same volume). Step 2: Take log₁₀ (common logarithm, NOT natural log) of the ratio. Step 3: Add to pKa. Example: acetate buffer (pKa 4.76) with [acetate] = 0.15 M and [acetic acid] = 0.10 M. Ratio = 0.15/0.10 = 1.5. log₁₀(1.5) = 0.176. pH = 4.76 + 0.176 = 4.94. Check: pH > pKa because more base than acid (✓). This method works for any buffer system once you know pKa and concentrations or ratio.
What happens to buffer pH when I add strong acid or base?
Adding strong acid (H⁺) or base (OH⁻) to a buffer triggers a neutralization reaction that changes the ratio [A⁻]/[HA], causing a small pH shift (much smaller than unbuffered). Adding H⁺: it reacts with conjugate base (A⁻ + H⁺ → HA), decreasing [A⁻] and increasing [HA] by the amount of H⁺ added. Ratio decreases, so pH drops (but only slightly). Adding OH⁻: it reacts with weak acid (HA + OH⁻ → A⁻ + H₂O), decreasing [HA] and increasing [A⁻]. Ratio increases, so pH rises (but only slightly). The buffer 'absorbs' the addition through equilibrium shifts, preventing large pH swings. To calculate new pH: (1) update [A⁻] and [HA] using stoichiometry (1:1 reaction), (2) calculate new ratio, (3) apply Henderson–Hasselbalch with new ratio.
Why can't I use a strong acid/base to make a buffer?
Buffers require a weak acid/conjugate base (or weak base/conjugate acid) pair in equilibrium. Strong acids (HCl, H₂SO₄, HNO₃) and strong bases (NaOH, KOH) dissociate completely—they don't establish equilibrium, so there's no weak acid or conjugate base to absorb pH changes. Example: mixing HCl with NaCl doesn't create a buffer; HCl dissociates fully (HCl → H⁺ + Cl⁻), and Cl⁻ is such a weak base (conjugate of strong acid) that it doesn't react with H⁺. Adding base would just neutralize all the HCl instantly, causing a huge pH jump. Only weak acids (like CH₃COOH, H₃PO₄⁻, NH₄⁺) partially dissociate, maintaining equilibrium between HA and A⁻—this equilibrium is what enables buffering via Le Chatelier shifts.
What is buffer capacity and how does it relate to concentration?
Buffer capacity (β) quantifies how many moles of strong acid or base can be added per liter to shift pH by 1 unit. It's maximized when pH = pKa (equal [A⁻] and [HA]) and increases with total buffer concentration. Formula: β ≈ 2.303 × [HA] × [A⁻] / ([HA] + [A⁻]). At pH = pKa, this simplifies to β ≈ 0.576 × C_total. Practical meaning: a 0.1 M buffer at pH = pKa has β ≈ 0.058 mol/(L·pH)—adding 0.058 mol/L of acid/base shifts pH by 1 unit. A 0.01 M buffer (10× more dilute) has β ≈ 0.0058 (10× less capacity). Higher concentration = higher capacity = better resistance to pH changes. Biochemists use 50-100 mM buffers for experiments because high capacity ensures pH stability despite metabolic acids/bases.
How do I find the required ratio [A⁻]/[HA] for a specific target pH?
Rearrange Henderson–Hasselbalch to solve for ratio: [A⁻]/[HA] = 10^(pH - pKa). Step 1: Subtract pKa from target pH to get (pH - pKa). Step 2: Raise 10 to this power (use 10^x button on calculator). Step 3: This is your required ratio. Example: you want pH 7.4 using phosphate buffer (pKa 7.2). pH - pKa = 7.4 - 7.2 = 0.2. Ratio = 10^0.2 ≈ 1.58. Meaning: you need 1.58× as much HPO₄²⁻ (base) as H₂PO₄⁻ (acid). For example, 0.158 M HPO₄²⁻ and 0.100 M H₂PO₄⁻. This method works for any buffer system—just ensure target pH is within pKa ± 1 for effective buffering.
Can I use Henderson–Hasselbalch at any pH or temperature?
Henderson–Hasselbalch is an approximation that works well under typical student problem conditions: (1) pH between 3 and 11, (2) buffer concentrations 0.01-1 M, (3) dilute aqueous solutions where activities ≈ concentrations, (4) temperature near 25°C (adjust pKa if specified otherwise). It assumes: complete dissociation of salts (e.g., NaCH₃COO → Na⁺ + CH₃COO⁻), no side reactions, and that [H⁺] from water auto-ionization is negligible compared to buffer [H⁺]. At extreme pH (&lt; 3 or &gt; 11), very dilute buffers (&lt; 0.001 M), or high ionic strength, assumptions break down and exact equilibrium calculations or activity corrections are needed. For homework, exams, and conceptual learning, HH is perfectly appropriate and gives accurate answers.
What are some common buffer systems and when should I use them?
Acetate (CH₃COOH/CH₃COO⁻, pKa 4.76): pH 3.8-5.8, enzyme assays, protein electrophoresis. Phosphate (H₂PO₄⁻/HPO₄²⁻, pKa₂ 7.2): pH 6.2-8.2, versatile for near-neutral biological work, HPLC. Tris (Tris-HCl, pKa 8.06 at 25°C): pH 7-9, standard in molecular biology (DNA/RNA work, PCR, Western blots). HEPES (pKa 7.5): pH 6.8-8.2, cell culture (Good's buffer, minimal metal chelation, low toxicity). Citrate (multiprotic, pKa 3.1/4.8/6.4): pH 2-6, food chemistry, acidic extractions. Ammonia (NH₃/NH₄⁺, pKa 9.25): pH 8.25-10.25, basic pH work. Carbonate (H₂CO₃/HCO₃⁻, pKa 6.4): physiological buffering in blood. Choose based on target pH (match pKa ± 1) and application constraints (e.g., Tris for DNA, HEPES for cells).
How does temperature affect buffer pH and pKa?
Temperature affects pKa of buffer systems through the van't Hoff equation (ΔH°/RT relationship). Most acid dissociations are endothermic, so pKa decreases with increasing temperature (acid gets stronger). Tris buffer has strong temperature dependence: dpKa/dT ≈ -0.03 per °C. If Tris pKa = 8.06 at 25°C, at 4°C it's ~8.7, and at 37°C it's ~7.7—a >1 pH unit range! Phosphate is more stable: dpKa/dT ≈ -0.003 per °C. If you prepare a buffer at 25°C and use it at 37°C without adjusting, pH will shift. Best practice: prepare and adjust buffers at their working temperature. For homework, assume 25°C unless stated otherwise. In real lab work, pH meters must be temperature-compensated, and some protocols specify 'pH at 25°C' or 'pH at 37°C.'
Why do my homework problems give me pH and ask for ratio instead of the reverse?
This tests your ability to rearrange Henderson–Hasselbalch and understand the inverse relationship between pH/pKa and ratio. It's also how you'd approach buffer design: you know the pH you want (given by experiment requirements) and need to figure out what ratio of acid/base to mix. To solve: rearrange pH = pKa + log₁₀(ratio) to get log₁₀(ratio) = pH - pKa, then ratio = 10^(pH - pKa). This is the same equation, just solving for a different variable. Example problem: 'A buffer at pH 5.0 uses acetic acid (pKa 4.76). Find [acetate]/[acetic acid].' Answer: ratio = 10^(5.0 - 4.76) = 10^0.24 ≈ 1.74. You'd mix 1.74 times as much acetate as acetic acid. Mastering rearrangements builds algebraic fluency and prepares you for varied problem types.

Explore More Chemistry Calculators

Master buffer systems, acid-base equilibria, and quantitative chemistry with our comprehensive calculator suite

Browse All Chemistry Tools

How helpful was this calculator?

Buffer Maker (Henderson–Hasselbalch) | EverydayBudd | EverydayBudd